Solve the inequality
\[\dfrac{x+1}{x+2}>\dfrac{3x+4}{2x+9}.\]
Solution: From the given inequality,
\[\frac{x + 1}{x + 2} - \frac{3x + 4}{2x + 9} > 0,\]which simplifies to
\[-\frac{x^2 - x - 1}{(x + 2)(2x + 9)} > 0,\]or
\[\frac{x^2 - x - 1}{(x + 2)(2x + 9)} < 0.\]The solutions to $x^2 - x - 1 = 0$ are $x = \frac{1 \pm \sqrt{5}}{2}.$  We can fill in a sign chart as follows:

\[
\begin{array}{c|ccccc}
& x < -\frac{9}{2} & -\frac{9}{2} < x < -2 & -2 < x < \frac{1 - \sqrt{5}}{2} & \frac{1 - \sqrt{5}}{2} < x < \frac{1 + \sqrt{5}}{2} & \frac{1 + \sqrt{5}}{2} < x \\ \hline
2x + 9 & - & + & + & + & + \\
x + 2 & - & - & + & + & + \\
x - \frac{1 - \sqrt{5}}{2} & - & - & - & + & + \\
x - \frac{1 + \sqrt{5}}{2} & - & - & - & - & + \\
\frac{x^2 - x - 1}{(x + 2)(2x + 9)} & + & - & + & - & +
\end{array}
\]Thus, the solution to $\frac{x^2 - x - 1}{(x + 2)(2x + 9)} < 0$ is
\[x \in \boxed{\left( -\frac{9}{2} , -2 \right) \cup \left( \frac{1 - \sqrt{5}}{2}, \frac{1 + \sqrt{5}}{2} \right)}.\]